HDU - 6097 Mindis(反演变换)

题目大意：

分析：

未AC代码：

#include<bits/stdc++.h>

using namespace std;

double r,px,py,qx,qy,ax,ay,bx,by;
int test;

int main()
{
scanf("%d",&test);
while(test--)
{
scanf("%lf%lf%lf%lf%lf",&r,&px,&py,&qx,&qy);
double op,oq,pq;
op=sqrt(px*px+py*py);
oq=op;
pq=sqrt((px-qx)*(px-qx)+(py-qy)*(py-qy));
if(op+oq-pq<1e-9)
{
double ans=2.0*sqrt(op*op+r*r);
printf("%.7f\n",ans);
continue;
}
double oa=r*r/op;
double c=(op*op+oq*oq-pq*pq)/(2.0*op*oq);
double cc=sqrt((c+1)/2.0);
//cout<<cc<<endl;
/*if(cc<=1e-6)
{
double ans=oa/r/pq;
//cout<<1;
printf("%lf\n",ans);
continue;
}*/
double dis=oa*cc;
if(dis>r)
{
double x=sqrt(op*op+r*r-cc*2.0*op*r);
double ans=2*x;
printf("%.7f\n",ans);
//cout<<"dsa";
}
else
{
double ans=pq*r/op;
printf("%.7f\n",ans);
}-
}

}